Question

Let \(f(x)=x^{3}+x\) Evaluate \(\lim _{s \rightarrow 0} \frac{f(x+s)-f(x)}{s}\).

Short answer

The limit is \(3x^2 + 1\).

Step-by-step solution

Substitute f(x+s) into the Difference Quotient

First, we substitute the expression for \(f(x)\) into the difference quotient:\[\lim_{s \to 0} \frac{f(x+s)-f(x)}{s} = \lim_{s \to 0} \frac{((x+s)^3 + (x+s)) - (x^3 + x)}{s}\].

Expand the Cubic Term

Expand the term \((x+s)^3\) using the binomial theorem:\[(x+s)^3 = x^3 + 3x^2s + 3xs^2 + s^3\].

Substitute the Expanded Term Back

Substitute the expanded form into the limit expression:\[\lim_{s \to 0} \frac{(x^3 + 3x^2s + 3xs^2 + s^3 + x + s) - (x^3 + x)}{s}\].

Simplify the Numerator

Cancel out \(x^3\) and \(x\) from the numerator:\[\lim_{s \to 0} \frac{3x^2s + 3xs^2 + s^3 + s}{s}\].

Factor Out the Common Term

Factor \(s\) from the numerator:\[\lim_{s \to 0} \frac{s(3x^2 + 3xs + s^2 + 1)}{s}\].The \(s\) in the numerator and the denominator cancel out.

Evaluate the Limit

After canceling \(s\), evaluate the limit as \(s\) approaches 0:\[\lim_{s \to 0} (3x^2 + 3xs + s^2 + 1) = 3x^2 + 1\].

Key concepts

Limit of a function

The limit of a function refers to the value that a function approaches as the input approaches a particular point. This concept is crucial in calculus, especially when dealing with derivatives. In our exercise, we deal with the limit as \( s \) approaches 0 in the expression \( \lim_{s \to 0} \frac{f(x+s) - f(x)}{s} \).

This expression is part of what is known as the "difference quotient," used to define the derivative. Understanding limits also requires understanding that they are about approaching a value, not necessarily reaching it. It's like zooming in very closely on the graph of a function to see what value it is getting closer to as \( s \) becomes extremely small.

Practically:

• We investigate how the function behaves near a point.
• Calculating the limit helps uncover the function's tangent at that point.

The elegance of limits lies in capturing the idea of instantaneous change, which is the essence of derivatives.

Difference quotient

The difference quotient is a formula that provides the average rate of change of a function over a specific interval. It is represented by \( \frac{f(x+s) - f(x)}{s} \) in our exercise.

This quotient is used to compute the derivative of a function at a point and is the core of many calculus problems. By understanding the difference quotient, you can better interpret how a function changes as its input changes. It essentially tells us how steep the tangent line is at any given point on the curve of the function.

Key points to remember:

• The difference quotient represents the slope of the secant line joining two points on a function's graph.
• By taking the limit as \( s \) approaches 0, the secant line becomes the tangent line, providing the slope of the curve at that specific point.

Grasping this concept is fundamental for understanding derivatives and is a stepping stone to more advanced calculus topics.

Binomial theorem

The binomial theorem is a powerful tool in algebra that allows you to expand expressions raised to a power, such as \((x+s)^3\) in our problem.

In the step-by-step solution, the binomial theorem helps expand \((x+s)^3\) to \(x^3 + 3x^2s + 3xs^2 + s^3\).

The general formula for expanding \((a+b)^n\) is:

• \((a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

This means you sum up products of binomial coefficients and powers of each term. Using the binomial theorem:

• Simplifies the complex operation of expansion.
• Allows calculations that are otherwise manual and lengthy.

Understanding it not only simplifies calculations but also strengthens your base in algebraic manipulations.

Simplification

Simplification is the process of reducing expressions into their simplest form, making them easier to work with and interpret.

In the solution, simplification involves canceling out terms in the numerator of the fraction to make computations clearer and more straightforward. Once the expansion is complete, terms like \(x^3\) and \(x\) cancel themselves out, leaving the simplified form \( \frac{3x^2s + 3xs^2 + s^3 + s}{s} \).

After this, factor out the common factor, \(s\), in the numerator:

• This step is crucial because it allows us to cancel \(s\) from both the numerator and the denominator.
• Post-cancellation, evaluating the limit becomes much simpler.

Simplification helps transform a complex algebraic expression into something more manageable and direct, making further calculations quicker and more reliable.